A body cools from `60^(@)` C to `50^(@)` C in 6 minutes, the temperature of the surroundings being `25^(@)` C. What will be its temperature after another 6 minutes ?

Explanation

To solve the problem, we will use Newton's Law of Cooling, which states that the rate of loss of temperature of a body is directly proportional to the temperature difference between the body and its surroundings.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Initial temperature of the body, \( T_1 = 60^\circ C \)
- Final temperature of the body after 6 minutes, \( T_2 = 50^\circ C \)
- Surrounding temperature, \( T_s = 25^\circ C \)
- Time interval, \( t = 6 \) minutes

2. **Calculate the Temperature Change**:
- The change in temperature, \( \Delta T = T_1 - T_2 = 60 - 50 = 10^\circ C \)

3. **Set Up the Equation Using Newton's Law of Cooling**:
- According to Newton's Law, we can express the rate of cooling as:
\[ \frac{\Delta T}{t} = k \cdot (T_{avg} - T_s) \]
- Here, \( T_{avg} \) is the average temperature during the cooling period, which can be calculated as:
\[ T_{avg} = \frac{T_1 + T_2}{2} = \frac{60 + 50}{2} = 55^\circ C \]

4. **Substitute Values into the Equation**:
- The equation becomes:
\[ \frac{10}{6} = k \cdot (55 - 25) \]
- Simplifying gives:
\[ \frac{10}{6} = k \cdot 30 \]

5. **Solve for the Constant \( k \)**:
- Rearranging gives:
\[ k = \frac{10}{6 \cdot 30} = \frac{1}{18} \]

6. **Calculate the Temperature After Another 6 Minutes**:
- Now, we need to find the temperature \( T \) after another 6 minutes, starting from \( T_2 = 50^\circ C \).
- Set up the equation again:
\[ \frac{50 - T}{6} = k \cdot \left( \frac{50 + T}{2} - 25 \right) \]
- Substituting \( k = \frac{1}{18} \):
\[ \frac{50 - T}{6} = \frac{1}{18} \cdot \left( \frac{50 + T}{2} - 25 \right) \]

7. **Simplify the Equation**:
- The right side simplifies to:
\[ \frac{50 + T}{2} - 25 = \frac{T}{2} \]
- Thus, the equation becomes:
\[ \frac{50 - T}{6} = \frac{1}{18} \cdot \frac{T}{2} \]

8. **Cross Multiply to Solve for \( T \)**:
- Cross multiplying gives:
\[ 18(50 - T) = 3T \]
- Expanding and rearranging:
\[ 900 - 18T = 3T \implies 900 = 21T \implies T = \frac{900}{21} = \frac{300}{7} \approx 42.86^\circ C \]

9. **Final Result**:
- The temperature of the body after another 6 minutes is approximately \( 42.86^\circ C \).